A mass m = 0.068 kg of benzene vapor (Lv = 3.94x105 J/kg) at its boiling point of 80.1°C is to be condensed by mixing with water at 50.0°C. What is the minimum mass of water required to condense all of the benzene vapor? Assume the mixing and condensation take place is a perfectly insulating container.

Answer 1

mL= =m₁c ΔT
m₁ = mL/c ΔT =
=0.068•3.94•10⁵/4183•30.1=
=0.213 kg